Q: Prove $\log_{x+5}^25 \cdot \log_5(x^2-x-6)=1$ Prove $\log_{x+5}^25 \cdot \log_5(x^2-x-6)=1$
Basically, I wanted to write log base $\mod{x+5}^2$ 5 but I wasn't able to write it. If anyone can just let me know how to do that I will be really grateful. Also I wanted to know Why mod(x+5}^2 will not be equal to 1. How to solve this problem?
Please don't downvote it guys.

A: I will assume the problem should state
$$\left(\log_{(x+5)^2}{5}\right)\cdot\left(\log_5{(x^2-x-6)}\right)=1.$$
To solve this, a common property of logs is that
$$\log_a{b}=\frac{1}{\log_b{a}}.$$
Can you take it from here?

A: The domain is $(x+5)^2>0$, $(x+5)^2\neq1$ and $x^2-x-6>0$.
Hence, we need to solve
$$\log_5(x^2-x-6)=\log_5(x+5)^2$$ or
$$x^2-x-6=x^2+10x+25$$ or
$$x=-\frac{31}{11}$$
and checking of the domain gives the answer: $\{-\frac{31}{11}\}$.
Now, why $(x+5)^2\neq1$.
If we write $\log_ab$ then by the definition of $\log$ we need $a>0$, $a\neq1$ and $b>0$.
But let we wrote $\log_12$ and let $\log_12=a$. 
Thus, $1^a=2$ and what we'll do with this thing?
